$$ S_n = 1+2+\cdots+n=\frac{n(n+1)}{2}.$$ How do I go about solving this ? I tried to search for a solution, but I didn't find anything ( well I didn't know what this was called to even look for it in the first place).
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Gauss solved this! – Dr. Sonnhard Graubner Aug 19 '17 at 15:30
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Try searching "Induction". – Donald Splutterwit Aug 19 '17 at 15:31
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okay thanks i'll try looking into those – Dahen Aug 19 '17 at 15:32
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Search tip: https://www.google.com/search?q=site%3Amath.stackexchange.com+sum+of+integers+n – Hans Lundmark Aug 19 '17 at 15:34
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@HansLundmark yeah that one is it, i'll check it out now – Dahen Aug 19 '17 at 15:34
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Incase you are a visual or tactile learner: you can draw a triangular "stair case" that has 1 step, then 2 steps, then 3 .... then $n$. Then draw another of these. Turn one upside down and put it on top of the other... and you have a rectangle made up $n$ by $n+1$ steps. https://latex.artofproblemsolving.com/4/2/3/4234a9f1ce8beb18aab9f27831bf936117909db1.png – fleablood Aug 19 '17 at 15:46
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@fleablood Oh! thanks that really makes it alot easier to understand – Dahen Aug 19 '17 at 15:47
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Another way to think of it (these are all the exact same proofs; just different ways of looking at them). Consider what the average of 1, 2,3,4,5.....n is. Is they are evenly distributed the average is right smack dab in the middle: $\frac {n+1}2$. And there are $n$ terms so they add up to $n*\frac {n+1}2$. – fleablood Aug 19 '17 at 21:04
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HINT: $$1+2+3+4+...+97+98+99+100=(1+100)+(2+99)+(3+98)+...$$ Can you solve this? the reult is $5050$ and now you can try it with variables!
Dr. Sonnhard Graubner
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Write $S:=1+2+\cdots+n$ and consider $$ \begin{matrix} S & = & 1 & + & 2 & + & \cdots & + & (n-1) & + & n \\ S & = & n & + & (n-1) & + & \cdots & + & 2 & + & 1 \\ 2S & = & (n+1) & + & (n+1) & + & \cdots & + & (n+1) & + & (n+1). \end{matrix} $$ Then $2S=n(n+1)$, which implies $S=n(n+1)/2$.
John Griffin
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why did you write $(n-1)$ ? thats the part I'm having difficulty understanding edit: never mind i just realized it, man the sleepiness is getting to my head apparently – Dahen Aug 19 '17 at 15:37
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@user464154 I'm just writing the sum $1+2+\cdots+(n-1)+n$ backwards, which is $n+(n-1)+\cdots+2+1$. – John Griffin Aug 19 '17 at 15:39
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$$2S_n=(1+n)+(2+n-1)+...+(n+1)=n(n+1),$$ which gives $$S_n=\frac{n(n+1)}{2}.$$
Michael Rozenberg
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Look at the square of side length $s=n $. Divide it diagonally in half. That gives $\frac {n^2}2$. Now go along the diagonal and "fill in " the missing half of each $1×1$ square, for $n\cdot \frac12$. Add the two to get $1+2+\dots+n $.